# Is the Archimedean principle necessary to prove the density of $\mathbb Q$ in $\mathbb R$?

I've noticed that most proofs of the density of $\mathbb Q$ in $\mathbb R$ use the Archimedean principle. For example see @Arturo Magidin's checked answer here. Density of irrationals

I'm puzzled by the fact that in the hyperreals, the standard rationals must be dense in the standard reals; yet the hyperreals are not Archimedean. This would seem to imply that the Archimedean principle is not a necessary assumption.

I can see that in the hyperreals, if $\epsilon$ is infinitesimal then there is no standard rational between $\epsilon$ and $2 \epsilon$; so that the standard rationals are not dense in the hyperreals.

But now I'm confused. Wouldn't the hyper-rationals be dense in the hyperreals? And wouldn't it still be a theorem that the standard rationals are dense in the standard reals?

Clearly the answer must be in some subtlety involving standard/nonstandard reals and the transfer principle. Can anyone shed any light? Is the Archimedean principle necessary to prove that the rationals are dense in the reals? Why aren't the hyperreals a counterexample?

• The rationals and reals are both Archimedean, regardless of whether you embed them in the hyperreals. How would embedding then in the hyperreals affect that? – Arthur Mar 25 '18 at 21:18
• Well, it's certainly not true that the Archimedean principle is necessary for every proof of the form "X is dense in Y". Obviously, what is needed to prove such a statement depends on what X and Y are... – Eric Wofsey Mar 25 '18 at 21:25
• Archimedean property $\neq$ Archimedean principle, which needs water. – Jean Marie Mar 25 '18 at 21:25
• @JeanMarie Eureka! – user4894 Mar 25 '18 at 21:27
• There is a direct proof that if we define the reals as the set of Dedekind cuts of rationals, modulo equivalence of cuts, then the rationals embed as a dense subset of the reals - no use of the Archimedean property in that proof. – Carl Mummert Mar 25 '18 at 21:30

I am not completely sure about the question, but I think this may be the answer.

When we construct the reals directly from the rationals, for example using Dedekind cuts or Cauchy sequences, we can directly prove the rationals are dense in the reals using the construction. Thus we can prove that the reals have the Archimedean property, we don't just assume they have the property.

However, sometimes authors want to avoid the details of this construction. They want to talk about the rationals and the reals, and the relationship between them, without having to talk about Dedekind cuts or anything like that. This way, they can move on to other topics more quickly - the constructions can be difficult for less experienced students. So, rather than constructing the reals from first principles, an author may present a list of axioms for the reals, so that those axioms are enough to obtain the results the author is interested in.

In these cases, it is common to use the Archimedean property itself as an axiom. This allows us to prove that the sequence $(1/n)$ converges to $0$, and other key facts about the topology of the reals, without referring to any specific construction of the reals.

Not every field is Archimedean, though. For example, the hyperreals are not Archimedean, as the question points out. There is no contradiction to this - when we actually construct the hyperreals we can't prove they are Archimedean (because they aren't), unlike when we construct the ordinary reals.

The same idea about replacing constructions with axioms applies to the hyperreals - some authors in nonstandard analysis give direct constructions of the hyperreals, while others only give a list of axioms for the hyperreals, which are enough to prove the results the authors are interested in obtaining.

• Yes that's exactly the answer to my question, thanks much. All the authors use the Archimedean property to prove the density of the rationals in the reals, but the Archimedean property is not necessary to the proof. Density can be proven directly from first principles (Dedekind cuts) without first proving the Archimedean property. That was my question. – user4894 Mar 25 '18 at 21:41
• Great - I'm glad that I was able to answer it for you! – Carl Mummert Mar 25 '18 at 21:42

As far as I can tell, your actual question is:

Proofs that the rationals are dense in the reals rely on the fact that the reals are Archimedean. But the hyperrationals are dense in the hyperreals, and the hyperreals are not Archimedean. How is this possible?

The obvious answer is that the hyperrationals are not the same as the rationals! If you wanted to prove the rationals were dense in the hyperreals, you would need the hyperreals to be Archimedean. But the hyperrationals are different from the rationals, so this is not relevant to proving that the hyperrationals are dense in the hyperreals.

## Completeness has many faces

For an ordered field $F$, the following are equivalent:

• $F$ has the Archimedean property and Cauchy sequences converge
• $F$ is Dedekind complete
• $F$ is connected
• $F$ has the LUB property
• $F$ has the monotone sequence property
• $F$ has the nested interval property
• In $F$, the Bolzano-Weierstrass theorem holds
• In $F$, the Heine-Borel theorem holds

There are probably other good examples too; this is the list appearing in Buck's Advanced Calculus.

Taking any one of these as an axiom suffices to be able to prove the rationals are dense in the reals.

## The Archimedean property is a necessary one

While we may have a proof that the rationals are dense in some field $F$ that doesn't use it, it must still be true that $F$ satisfies the Archimedean property; we can't get away from the fact it holds, even if we don't actually use it directly. Explicitly

Theorem: If $F$ is an ordered field in which the rationals are dense, then $F$ is Archimedean.

Proof: For any $x \in F$, there is a rational number $q$ with $|x| < q < |x| + 1$. $\lceil q \rceil$ is a natural number larger than $x$. $\square$

## The hyperreals are Archimedean too

Internally, the hyperreals are Archimedean. More precisely, if you take the Archimedean property

For every $r \in \mathbb{R}$ there is an $n \in \mathbb{N}$ such that $n > r$

and apply the transfer principle, you get a theorem about the hyperreals

For every $r \in {}^\star\mathbb{R}$ there is an $n \in {}^\star\mathbb{N}$ such that $n > r$

• That's an interesting point. That internally the hyperreals are Archimedean. A little beyond my understanding of the transfer principle, which is possibly the source of my confusion. I also understand now why Carl asked what I mean by necessary. I meant necessary to the proof, not necessarily true about the reals. The Archimedean property is necessarily true about the reals, but not necessary for the proof, since we can prove it from a construction of the reals. – user4894 Mar 25 '18 at 22:40
• I believe the headline "The hyperreals are Archimedean too" may lead to confusion. The hyperreals are simply not Archimedean. It is true that even the statement "The hyperreals are Achimedean" can't be stated in the internal language, and that the transfer principle shows that the hyperrationals dense in the hyperreals, but those are not what "The hyperreals are Archimedean" is defined to say. It is meaningful to ask "Are the hyperreals Archimedean", and the literal answer is that they aren't. – Carl Mummert Mar 26 '18 at 11:12

The truncation $10^{-n}\lfloor10^n x\rfloor$ of a real $x$ at rank $n$ of its decimal expansion is rational and implies that the rationals are dense in the reals. This argument works also in the hyperreals and does not use the Archimedean property.